Downtime Puzzle, again

[GM-Ayu]

Because you totally didn't see this coming!

For those wondering why server is down: this



Please post your detailed answer directly in this thread. I'll only accept the first correct, *UNEDITED* answer. I'll only accept the first post from a forum member as their answer, so please be careful and answer carefully without the need to edit (and of course, don't make a mistake in your only chance of answering.) In your solution post, also provide your in-game character names. Prizes will only be distributed to the mentioned IGN of the answer post.

Prizes are probably not "super charming" but it's more of a time burner for those who are bored or suffering heRO withdrawl symptoms.

If you got questions or just want to make comments, feel free to do so and of course you won't be counted by the "1 answer no edit" rule. Just that if you are answering, do so with a separate post, and put the answer in your comment/question post.


List of Winners/Prizes:

Ellie: 5 rainbow cake, 5 gb
Usopp: 3 GB
Mace: OCA
Hann: OCA


Question 1
Prize: 3 GB, 5 Rainbow Cake, 5 GB

"heRO GMs captured 7 members who are suspected to be participating in heRO rule breaking activities. They are identified as Suspect A, B, C, D, E, F and G. GMs interrogated the seven, and got the following results:

A admits to have met the other 6
B admits to have met 5 others
C admits to have met 4 others
D admits to have met 3 others
E admits to have met 2 others
F admits to have met 2 others
G admits to have met 1 others

None of them would have claimed to meet with more people than s/he has actually meet, because @nukes are painful.

How will the GMs know if they are telling the truth? Explain the reasoning why it is truthful, or explain the reasoning why someone is definitely lying."

[Ellie]

They are lying.

A would have spoken to them all, since there are only six others. Since A spoke to them all, it means G knows A only.
So we have..

A bcdefg
G a

This means the one person B did not speak to would be G, as G only knew A.
So we have..

A bcdefg
B acdef
G a

This means E and F must have already met A and B.

A bcdefg
B acdef
E ab
F ab
G a

D knows 3. Already it knows a and b, and cannot know e, f or g, so must know c.

A bcdefg
B acdef
D abc
E ab
F ab
G a

This leaves C, who apparently knows 4 others. However, it is only possible for him to know 3 others to fit in with the story. a, b and d.

[Aaronock]

Ah, someone in this group is a huge fat liar!
My reasoning is that each of them claim to know so many, but it doesn't add up when put in a chart.

A: Knows B, C, D, E, F, G
B: Knows A, C, D, E, F
C: Knows supposedly 4 people but only 3 add up
D: Knows A, B, C
E: Knows A, B
F: Knows A, B
G: Knows A

SO, C, E, or F are lying about how many of the people that they know.
I suck at riddles I probably only got this half right

[Captain Usopp]

dang was to slow I guess, I am going to actually try and work up a mathematical proof

[GM-Ayu]

Ellie got it correct!

Though my solution is much shorter:

"Every meeting must produce 2 acquaintanceships. If they are telling the truth, then there is an odd number of acquaintanceships. This is a contradiction."


(Question 2 is "in the process of being made" give me a moment)

[Galt]

-just noticed this thread-

Nice answer, Ellie. :D

[Aaronock]

Congrats Inasad, I'll hopefully win the next one! I almost had it damn me :O

[Galt]

Ellie's a sharp one, that she is.

[Ellie]

\o/

[SeifuSnow]

Err. Nice Ellie. xD -Deletes post-